Bondoc, Oliver Rostum II M.201413101 || BSCPT
Modified Questions
[19.50] The solubility of silver carbonate is 0.032 M at 20
degrees Celsius. Calculate its Ksp.
Modified Question:
The solubility of silver iodate is 0.89 M at 25 degrees Celsius.
Calculate its Ksp.
Solution:
AgIO3 Ag+ + IO3-x 10-3 M
RAgIO3 Ag+ + IO3-
I000
Cx 10-3 Mx 10-3 M
Ex 10-3 Mx 10-3 M
Ksp = [Ag+][IO3-] = (x 10-3 M)2 = 9.92 x 10-6
[19.51] The solubility of zinc oxalate is 7.9x10-3 at 18 degrees
Celsius. Calculate its Ksp.
Modified Question:
The solubility of calcium oxalate is 6.2x10-5 M at 25 degrees
Celsius. Calculate its Ksp.
Solution:
CaC2O4 Ca2+ + C2O4-2
R CaC2O4 Ca2+ + C2O4-2
I000
C
E
Ksp = [Ca2+][C2O4-2] = (x 10-7 M)2 = 2.34 x 10-13
[19.54] Find the molar solubility of SrCO3 (Ksp = 2.1x10-10) in
(a) pure water and (b) 0.13 M Sr(NO3)2.
Modifies Question: Find the molar solubility of NiCO3 (Ksp =
2.8x10-7) in (a) pure water and (b) 0.047 M Ni(NO3)2.
Solution:
(a) NiCO3 Ni2+ + CO32-
R NiCO3 Ni2+ + CO32-
I000
C
E
Ksp = [Ni2+][CO3-2] 2.8x10-7= ()2
x = 5.29 x 10-4
(b) NiCO3 Ni2+ + CO32-
R NiCO3 Ni2+ + CO32-
I00.047 M0
C
E
Ksp = [Ni2+][CO3-2]2.8x10-7= ()() x = 5.96 x 10-6
[19.55] Find the molar solubility of BaCrO4 (Ksp = 2.1x10-10) in
(a) pure water and (b) 1.5x10-3 M Na2CrO4.
Modified Question:
Find the molar solubility of PbCrO4 (Ksp = 3.7 x10-13) in (a)
pure water and (b) 1.6x10-3 M Na2CrO4.
Solution:
(a) PbCrO4 Pb2+ + CrO42-
R PbCrO4 Pb2+ + CrO42-
I000
C
E
Ksp = [Pb2+][CrO4-2] 3.7 x10-13= ()2 x = 6.08 x 10-7
(b) PbCrO4 Pb2+ + CrO42-R PbCrO4 Pb2+ + CrO42-
I001.6x10-3 M
C
E1.6x10-3 M +
Ksp = [Pb2+][CrO4-2]3.7 x10-13 = (1.6x10-3 M + )() x = 2.31 x
10-10
[20.27] For the reaction H2O(g) + Cl2O(g) 2HClO(g) , you know
Srxn and S of HClO(g) and H2O(g). Write an expression that can be
used to determine S of ClO2(g).
Modified Question:
For the reaction N2O(g) + NO2(g) 3NO(g) , you know Srxn and S of
NO(g) and N2O(g). Write an expression that can be used to determine
S of NO2(g).
Solution:
Srxn = Spdt - SrcntSrxn = 3(SNO) (S N2O + SNO2)Srxn = 3(SNO) S
N2O - SNO2
SNO2 = 3(SNO) S N2O - Srxn
[20.40] Calculate G for each reaction using Gf values(a) 2Mg(s)
+ O2(g) 2MgO(s)(b) 2CH3OH(g) + 3O2(g)2CO2(g) + 4H2O(g)
Modified Question:
Calculate G for each reaction using Gf values(a) H2(g) + I2(s)
2HI(g)(b) MnO2(s) + 2CO(g) Mn(s) + 2CO2(g)
Solution:
(a) G = 2(1.3) [(0) + (0)] = 2.6 (b) G = [2(-394.4) + 0]
[(-466.1) + 2(137.2)] = -597
[20.42] Find the G for the reaction in Problem 20.40 using Hf
and S values.
Modified Question:
Find the G for the reactions in the previous problem using Hf
and S values
Solutions:
(a) G = Hf - TS = {(25.9) [(0) + (0)]} (298K){ 2(0.20633)
[(0.1306) + (0.11614)]}G = - 2.4
(b) G = Hf - TS = {[2(-393.5) + 0] [(-520.9) + 2(110.5)]}
(298K){ [2(0.2137) 0.084] [(0.0531) + 2(0.1975)]}G = -596.4
[21.31] Balance each skeleton reaction, calculate Ecell and
state whether the reaction is spontaneous:
(a) Co(s) + H+(aq) Co2+(aq) + H2(g)(b) Mn2+(aq) + Br(l)
MnO4-(aq) + Br-(aq) [acidic]
Modified Question:
Balance each skeleton reaction, calculate Ecell and state
whether the reaction is spontaneous:
(a) Ag(s) + Cu2+(aq) Ag2+(aq) + Cu(s)(b) Mn2+(aq) + Co3+(aq)
MnO2(s) + Co2+(aq) [acidic]
Solutions:
(a) Ag(s) + Cu2+(aq) Ag2+(aq) + Cu(s) Ag(s)Ag2+(aq) + 2e-Ecell =
0.80 V 2e- + Cu2+(aq) Cu(s)Ecell = 0.34 V Ag(s) + Cu2+(aq) Ag2+(aq)
+ Cu(s)Ecell = 0.46 V spontaneous
[21.51] A voltaic cell consists of a standard reference
half-cell and a Cu/Cu2+ half-cell. Calculate [Cu2+] when Ecell is
0.22V
Modified Question:
A voltaic cell consists of a standard reference half-cell and a
Pb/Pb2+ half-cell. Calculate [Pb2+] when Ecell is 0.37V
Solution: Pb(s) Pb2+(aq) + 2e-
Ecell = Ecell - logQ0.37 V = -0.13 V - log [Pb2+][Pb2+] = 0.020
M
[21.52] A voltaic cell consists of an Mn/Mn2+ half-cell and a
Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 1.4 M and Ecell
is 0.44V.
Modified Question:
A voltaic cell consists of an Ag/Ag+ half-cell and a Co/Co2+
half-cell. Calculate [Co2+] when [Ag+] is 0.097 M and Ecell is
0.19V.
Solution:
Ag(s)Ag2+(aq) + 2e-Ecell = 0.80 V 2e- + Co2+(aq) Co(s)Ecell =
-0.28 V Ag2+(aq) + Co(s)Ag(s) + Co2+(aq)Ecell = 1.08 V
Ecell = Ecell - log Q0.19 V = 1.08 V - log
[Co2+] = 98.52 M
